$$\int_{\2}^{\9}(\ $\sqrt{\4+x^2}$ ) d\x $\$ $$

asked by guest
on Nov 18, 2024 at 6:51 am



You asked:

Evaluate the integral: ∫294+x2dx\int_{2}^{9}{\sqrt{4 + {x}^{2}}}dx

MathBot Answer:

Definite integral

∫294+x2dx=−22−2log⁡(1+2)+2arsinh⁡(92)+9852≈41.3154711762633853023529911313652\int_{2}^{9}{\sqrt{4 + {x}^{2}}}dx = - 2 \sqrt{2} - 2 \log{\left(1 + \sqrt{2} \right)} + 2 \operatorname{arsinh}{\left(\frac{9}{2} \right)} + \frac{9 \sqrt{85}}{2} \approx 41.3154711762633853023529911313652

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